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vodka [1.7K]
2 years ago
15

A 2000 N net force will give a car with some amount of mass an acceleration of 4 m/s2

Physics
1 answer:
hammer [34]2 years ago
3 0

Answer:

2m/s²

Explanation:

Given parameters:

Net force on the car  = 2000N

Acceleration = 4m/s²

Unknown:

Acceleration of a car twice the mass  = ?

Solution:

Let us first find the mass of the car;

 Force  = mass x acceleration

         Mass = \frac{Force }{acceleration}  

 Mass = \frac{2000}{4}   = 500kg

 Now,

     whose mass is twice that of the car

    Mass of the new car = 2 x 500  = 1000kg

So;

  Acceleration  = \frac{Net force }{mass}  

  Acceleration  = \frac{2000}{1000}  = 2m/s²

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A 912-kg car is being driven down a straight, level road at a constant speed of 31.5 m/s. When the driver sees a police cruiser
sergejj [24]

Answer:

786.6 N

Explanation:

mass of car, m = 912 kg

initial velocity of car, u = 31.5 m/s

final velocity of car, v = 24.6 m/ s

time, t = 8 s

Let a be the acceleration of the car

Use first equation of motion

v = u + a t

24.6 = 31.5 + a x 8

a = - 0.8625 m/s^2

Force, F = mass x acceleration

F = 912 x 0.8625

F = 786.6 N

Thus, the force on the car is 786.6 N.

8 0
2 years ago
An unknown immiscible liquid seeps into the bottom of an open oil tank. Some measurements indicate that the depth of the unknown
ira [324]

Answer:

The specific gravity of the unkown liquid is 15.

Explanation:

Gauge pressure, at the bottom of the tank in this case, can be calculated from

P_{bot}=\gamma_{oil}h_1 + \gamma_{unk}h_2,

where h_1 and h_2 are the height of the column of oil and the unkown liquid, respectively. Writing for \gamma_{unk}, we have

\gamma_{unk} = \frac{P_{bot} - \gamma_{oil}h_1}{h_2} = \frac{65\frac{kN}{m^2} - 8.5\frac{kN}{m^3}\times 5m}{1.5m} = \mathbf{15 \frac{kN}{m^3}}.

Relative to water, the unknow liquid specific weight is 15 times bigger, therefore this is its specific gravity as well.

3 0
2 years ago
What are the relative ages of the features in order of oldest to youngest?
yuradex [85]

Answer:

Layer 1, Rock 2, Rock 1, Fault

8 0
2 years ago
Dos cargas Q1=2pc y Q2=4pc estan separadas por una distancia de 6cm ¿con que fuerza se atraen?
noname [10]

Here we can use coulomb's law to find the force between two charges

As per coulombs law

]tex]F = \frac{kq_1q_2}{r^2}[/tex]

here we have

k = 9 * 10^9

q_1 = 2pC

q_2 = 4pC

r = 6cm = 0.06 m

now by using the above equation we have

F = \frac{9*10^9 * 2*10^{-12} * 4*10^{-12}}{0.06^2}

F = 2 * 10^{-11} N

so here the force between two charges is of above magnitude and this will be repulsive force between them as both charges are of same sign.

3 0
3 years ago
A giant armadilo moving northward with a constant acceleration covers the distance between two points 60m apart in 6 seconds. It
Naddika [18.5K]
Acceleration is the rate of change of the velocity of an object that is moving. This value is a result of all the forces that is acting on an object which is described by Newton's second law of motion. To determine acceleration, we need to know the initial velocity and the final velocity and the time elapsed. From the given values, we need t o calculate for the initial velocity. We use some kinematic equations. We do as follows:

 x = v0t + at^2/2
60 = v0(6) + a(6)^2/2
60 = 6v0 + 18a          (EQUATION 1)

vf = v0 + at
15 = v0 + a(6)
15 = v0 + 6a             (EQUATION 2)

Solving for v0 and a,
v0 = 5 m/s
a = 1.7 m/s^2
8 0
3 years ago
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